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Full text of "Treatise On Analysis Vol-Ii"

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We have seen in (15.12.4) that T(T)and J(T(T*)) are orthogonal supple-
ments of each other in E x E. Hence for each x e E there exists a unique
y e dom(T) and a unique z e dom(T*) such that

(                 (*, 0) - (y, T-y) + (T* - z, -z).

Put y = B - x and z = C - x. Clearly B and C are linear operators defined on the
whole of E, and we have B(E) a dom(T) and C(E) <= dom(T*). Also, by

W2 = \\y\\2 + \\T-y\\2 + Ni2 + nr*  z||2,

so that || 5  jt|| <| || jc|| and || C  x\\  \\x\\ . Hence B and C are continuous. The
relation ( is equivalent to

x = B-x+T*C-x      and       Q=-C-x + TB-x,

so that C=TB and T(B(E)) c dom(T*), hence J9(E) <= dom(T*T). Conse-
quently T* TB is defined on all of E, and we have

which shows that B is injective and /+ T*T surjective. For each
w e dom(T*T) we have

(           (w + T*T- w \ w) = |M|2 + (T*T- w \ w)

- \\w\\2 + \\T  w\\2

because T= T**; this shows that the relation w + T*T- w = 0 implies that
w = 0, and hence that /+ T*T is a bijective mapping of dom(T*r) onto E.
Also, since T(B) is closed in E x E (15.12.2), the same is true of F(/+ T*T)
(15.12.5), and it follows immediately (15.12.5) that T*Tis closed. We next
remark that, for all, u, v in E,,we have

(B-u\v) = (B-u\B'V+ T*TB  v)

T*T)B -u\B-v) = (u\B-v).

Hence B is.self-adjoint. Also, replacing w by B - x in ( we obtain, for
each x e E, e E such that